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### Messages - Xiaoning Han

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##### Quiz-7 / Re: TUT 0301
« on: November 30, 2018, 03:58:15 PM »
$$x^2+ix+2+i=0 \Rightarrow x=\frac{-\pm\sqrt{-9-4i}}{2}$$
$1)$.
$x\in[0,R]$,
\begin{align}f(x)&=x^2+ix+2+i \\&=(x^2+2)(1+\frac{i(x+1)}{x^2+2})\end{align} \\f(0)=2+i\\f(R)=R^2+iR+2+i
Thus, $\arg(f(x))$ starts from $\arctan(\frac{1}{2})$ and ends on $\arctan(\frac{R+1}{R^2+2})$.
$\\$
$2)$.
$x=Re^{it},0\leq t\leq \frac{\pi}{2}$,
\begin{align}f(Re^{it})&=(Re^{it})^2+i(Re^{it})+2+i \\&=R^2(e^{it}-\frac{ie^{it}}{R}+\frac{2+i}{R^2})\end{align}, 0\leq t \leq \pi.
$\\$
$3)$.
$x=iy, 0 \leq y \leq R$,
$$f(iy)=-y^2-y+2+i \\ y=0 \Rightarrow 2+i \\ y=R \Rightarrow -R^2-R+2+i$$
Thus, it is a horizontal line.
$\\$
Therefore, the net changes of the argument is $0$, there are no zeros in the first quadrant.

2
##### End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 3
« on: November 27, 2018, 01:39:23 PM »
Let $$f(z)=\frac{\cos(\frac{z}{6})+z\cdot\sin z}{\sin^2z}$$
$1)$
$\\$
Let $$g(z)=\cos(\frac{z}{6})+z\sin z,g(z)\neq 0,\therefore order =0.$$
Let $$h(z)=\sin^2 z, h(0)=0\\h'(z)=2\sin z\cos z=\sin 2z, h'(0)=0\\ h''(z)=2\cos 2z, h''z(0)\neq0, \therefore order=2.\\ 2-0=2, \therefore \text{ it is a pole of order } 2.$$

$2)$
$\\$
Let $$z=3\pi + 6k\pi\\g(3\pi + 6k\pi)=0 \\ g'(z)=-\frac{1}{6}\sin(\frac{z}{6}) +z\cos z\\ g'(3\pi + 6k\pi)\neq 0,\therefore order =1.$$
Let $$h(z)=\sin^2 z, h(3\pi + 6k\pi)=0\\h'(z)=\sin 2z, h'(3\pi + 6k\pi)=0\\h''(z)=2\cos 2z, h''(3\pi + 6k\pi)\neq0\therefore order=2.\\2-1=1, \text{ order 1 simple pole}.$$

$3)$
$\\$
$$z=k\pi (k\neq 0), z\neq 3\pi + 6k\pi$$
Let $$g(z)=\cos(\frac{z}{6})+z\sin z,g(k\pi\neq 0),order=1$$
Let $$h(z)=\sin^2 z, h(k\pi)=0\\h'(z)=\sin 2z,h'(k\pi)=0\\h''(z)=2\cos 2z,h''(k\pi)\neq0, \therefore order=2$$
$$2-1=1, \text{ it is a simple pole}.$$

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